Question

What is value of d2y / dx2 of x^2 + y^2 =25

Answer 1

Answer:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = -\dfrac{1}{y} - \dfrac{x^2}{y^3}}$

Step-by-step explanation:

We are given an equation of circle with radius of 5 units:

$\displaystyle{x^2+y^2=25}$

To find the second derivative, you'd have to differentiate the equation twice.

We can use implicit differentiation to differentiate. Its concept is to derive like a normal for both sides of equation but since we are differentiating with respect to x and we have y-term, we derive normally then multiply by dy/dx or y'.

For simple clarification, you derive normal and apply chain rules. Hence why there's dy/dx multiplied by 2y:

$\displaystyle{\dfrac{d}{dx}x^2 +\dfrac{d}{dx} y^2 = \dfrac{d}{dx}25}$

Recall the power rules:

$\displaystyle{\dfrac{d}{dx}ax^n =n\cdot ax^{n-1}}$

Chain Rules:

$\displaystyle{\dfrac{d}{dx}u^n = n u^{n-1} \cdot \dfrac{du}{dx}}$

Hence:

$\displaystyle{2x^{2-1} \cdot \dfrac{dx}{dx}+ 2y^{2-1} \cdot \dfrac{dy}{dx} = 0}$

Note that deriving a constant will always result in 0.

Simplify:

$\displaystyle{2x+ 2y \dfrac{dy}{dx} = 0}$

Solve for dy/dx:

$\displaystyle{2x+ 2y \dfrac{dy}{dx} = 0}\\\\\displaystyle{2y \dfrac{dy}{dx} = -2x}\\\\\displaystyle{\dfrac{dy}{dx} = \dfrac{-2x}{2y}}\\\\\displaystyle{\dfrac{dy}{dx} = -\dfrac{x}{y}}$

We've finally found the first derivative of relation. However, we have to find the second derivative, so we derive dy/dx to find the second derivative:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2}}$

Therefore:

$\displaystyle{\dfrac{d}{dx}\left(-\dfrac{x}{y}\right)}$

For this, we will be using quotient rules. Keep in mind that both x and y are function!

Recall quotient rules:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}}$

Let u = -x and v = y:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = \dfrac{(-x)'y - (-x)y'}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = \dfrac{-1\cdot y - (-x)\cdot \dfrac{dy}{dx}}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y+x\dfrac{dy}{dx}}{y^2}}$

Now we know that dy/dx = -x/y, so we substitute dy/dx as -x/y in the second derivative:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y+x\dfrac{dy}{dx}}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y+x\left(-\dfrac{x}{y}\right)}{y^2}}$

Simplify:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y-\dfrac{x^2}{y}}{y^2}}$

More simplification:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y}{y^2} - \dfrac{\dfrac{x^2}{y}}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = -\dfrac{1}{y} - \dfrac{x^2}{y^3}}$

Therefore, the second derivative is:

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = -\dfrac{1}{y} - \dfrac{x^2}{y^3}}$

Answer 2

Answer:

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - \frac{ {x}^{2} }{ \sqrt{ {(25 - {x}^{2} )}^{3} } }$

Step-by-step explanation:

$x^2 + y^2 = 25$

${y}^{2} = 25 - {x}^{2}$

$y = \sqrt{(25 - {x}^{2} )}$

We have to find the double derivative of above equation,

let's find out the first derivative of above equation,

$\frac{dy}{dx} = \frac{d}{dx} \sqrt{(25 - {x}^{2} )}$

We know that,

$\frac{d}{dx} ( \sqrt{x} ) = \frac{1}{2 \sqrt{x} }$

but one thing that we should keep in mind, the term inside the root is not x hence we will have to re-diffrentiate the term inside the root.

$\frac{d}{dx} \sqrt{(25 - {x}^{2} )}= \frac{1}{2\sqrt{(25 - {x}^{2} )}} \frac{d}{dx}{(25 - {x}^{2} )}$

Derivative of any constant number equals zero,

$\frac{d}{dx} \sqrt{(25 - {x}^{2} )} = \frac{1}{2\sqrt{(25 - {x}^{2} )}} ( - 2{x})$

Simplifying above result,

$\frac{dy}{dx} = \frac{ - {x}}{\sqrt{(25 - {x}^{2} )}}$

Now let's take the second derivative,

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} \frac{ - {x}}{\sqrt{(25 - {x}^{2} )}}$

we can write above term in the form of U.V of derivative

$\frac{d}{dx} U.V = U\frac{d}{dx}V + V\frac{d}{dx}U$

Similarly,

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} (x \cdot\frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} )$

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} \frac{d}{dx} x + x\frac{d}{dx}\frac{ - {1}}{\sqrt{(25 - {x}^{2} )}}$

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} + x\frac{d}{dx}\frac{ - {1}}{\sqrt{(25 - {x}^{2} )}}$

Now we know that,

$\frac{d}{dx} \frac{1}{ \sqrt{x} } = - \frac{1}{2 \sqrt{ {x}^{3} } }$

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - x \frac{-1}{2 \sqrt{ {(25 - {x}^{2} )}^{3} } } \frac{d}{dx} (25 - {x}^{2} )$

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - x\frac{-1}{2 \sqrt{ {(25 - {x}^{2} )}^{3} } } ( -2 {x})$

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - \frac{ {x}^{2} }{ \sqrt{ {(25 - {x}^{2} )}^{3} } }$

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